Lecture 3. Pythagorean Triples

Oleg Viro

February 1, 2016

1/11 Warm Up Exercise

Write down the Pythagorean Theorem.

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 .

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Write down the converse theorem.

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c .

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? •

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? • How to prove? •

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? • How to prove? • Is it possible that numbers a , b , c such that a2 + b2 = c2 are all • positive integers?

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? • How to prove? • Is it possible that numbers a , b , c such that a2 + b2 = c2 are all • positive integers? A triple (a,b,c) of positive integers such that a2 + b2 = c2 is called a Pythagorean triple.

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? • How to prove? • Is it possible that numbers a , b , c such that a2 + b2 = c2 are all • positive integers? A triple (a,b,c) of positive integers such that a2 + b2 = c2 is called a Pythagorean triple. A triangle with sides 3,4,5 is called the Egyptian triangle.

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? • How to prove? • Is it possible that numbers a , b , c such that a2 + b2 = c2 are all • positive integers? A triple (a,b,c) of positive integers such that a2 + b2 = c2 is called a Pythagorean triple. A triangle with sides 3,4,5 is called the Egyptian triangle. Why?

2/11 Warm Up Exercise

Write down the Pythagorean Theorem. Correct Answer. In any right triangle, the length c of its hypothenuse and lengths a , b of its legs, a2 + b2 = c2 . Converse Theorem. For any real positive numbers a , b , c such that a2 + b2 = c2 , there exists a right triangle with legs a , b , and hypotenuse c . Is this true? • How to prove? • Is it possible that numbers a , b , c such that a2 + b2 = c2 are all • positive integers? A triple (a,b,c) of positive integers such that a2 + b2 = c2 is called a Pythagorean triple. A triangle with sides 3,4,5 is called the Egyptian triangle. Why? Problem: Find all Pythagorean triples.

2/11 • Warm Up Exercise

Number theory approach Diophantine equation • Primitive solutions • Arithmetics of evens and odds • Squares modulo 4 • New unknowns • Solution • Done? Number theory approach • Pythagorian points • Parametrizations of the circle Diophantine equation

3/11 Primitive solutions

If (a,b,c) is a Pythagorean triple, then for any k N ∈ the triple (ka,kb,kc) is Pythagorean.

4/11 Primitive solutions

If (a,b,c) is a Pythagorean triple, then for any k N ∈ the triple (ka,kb,kc) is Pythagorean. A Pythagorean triple is said to be primitive if it cannot be obtained as (ka,kb,kc) with k N , k > 1 from a Pythagorean triple (a,b,c) . ∈

4/11 Primitive solutions

If (a,b,c) is a Pythagorean triple, then for any k N ∈ the triple (ka,kb,kc) is Pythagorean. A Pythagorean triple is said to be primitive if it cannot be obtained as (ka,kb,kc) with k N , k > 1 from a Pythagorean triple (a,b,c) . ∈ Lemma 1. A Pythagorean triple (a,b,c) is primitive if and only if a , b and c have no common divisors.

4/11 Primitive solutions

If (a,b,c) is a Pythagorean triple, then for any k N ∈ the triple (ka,kb,kc) is Pythagorean. A Pythagorean triple is said to be primitive if it cannot be obtained as (ka,kb,kc) with k N , k > 1 from a Pythagorean triple (a,b,c) . ∈ Lemma 1. A Pythagorean triple (a,b,c) is primitive if and only if a , b and c have no common divisors. Lemma 2. Presence of odds. In a primitive Pythagorean triple, at least one of the numbers is odd.

4/11 Arithmetics of evens and odds

Addition: even + even = even ,

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd ,

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

In equality X + Y = Z , the number of odd numbers is even (2 or 0),

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

In equality X + Y = Z , the number of odd numbers is even (2 or 0), the number of even numbers is odd (1 or 3).

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

In equality X + Y = Z , the number of odd numbers is even (2 or 0), the number of even numbers is odd (1 or 3).

Squaring: n N is odd n2 is odd. ∈ ⇐⇒

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

In equality X + Y = Z , the number of odd numbers is even (2 or 0), the number of even numbers is odd (1 or 3).

Squaring: n N is odd n2 is odd. ∈ ⇐⇒ In a primitive Pythagorean triple, two numbers are odd and one is even.

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

In equality X + Y = Z , the number of odd numbers is even (2 or 0), the number of even numbers is odd (1 or 3).

Squaring: n N is odd n2 is odd. ∈ ⇐⇒ In a primitive Pythagorean triple, two numbers are odd and one is even.

Which numbers can be odd? a , b , c ?

5/11 Arithmetics of evens and odds

Addition: even + even = even , even + odd = odd , odd + odd = even .

In equality X + Y = Z , the number of odd numbers is even (2 or 0), the number of even numbers is odd (1 or 3).

Squaring: n N is odd n2 is odd. ∈ ⇐⇒ In a primitive Pythagorean triple, two numbers are odd and one is even.

Which numbers can be odd? a , b , c ?

The answer: c must be odd, one of a and b is even, another one is odd.

5/11 Squares modulo 4

The square of an even number is divisible by four.

6/11 Squares modulo 4

The square of an even number is divisible by four.

The square of an odd number has remainder 1 under division by 4 .

6/11 Squares modulo 4

The square of an even number is divisible by four.

The square of an odd number has remainder 1 under division by 4 .

Notation: x y mod m means x y is divisible by m . ≡ −

6/11 Squares modulo 4

The square of an even number is divisible by four.

The square of an odd number has remainder 1 under division by 4 .

Notation: x y mod m means x y is divisible by m . ≡ − We say: x is congruent to y modulo m .

6/11 Squares modulo 4

The square of an even number is divisible by four.

The square of an odd number has remainder 1 under division by 4 .

Notation: x y mod m means x y is divisible by m . ≡ − We say: x is congruent to y modulo m . Introduced in 1798 by Carl Friedrich Gauss (Gauβ) (1777-1855), when he was 21, in Disquisitiones Arithmeticae.

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 .

Notation: x y mod m means x y is divisible by m . ≡ −

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ −

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Exercise. Prove that moreover n 1 mod2 = n2 1 mod8 . ≡ ⇒ ≡

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Theorem. If c is an integer such that c 2 , or 3 mod4 , ≡2 then the equation x = c has no integer solution.

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Theorem. If c is an integer such that c 2 , or 3 mod4 , ≡2 then the equation x = c has no integer solution. If both a and b are odd, then a2 + b2 1+1 2 mod4 . ≡ ≡

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Theorem. If c is an integer such that c 2 , or 3 mod4 , ≡2 then the equation x = c has no integer solution. If both a and b are odd, then a2 + b2 1+1 2 mod4 . ≡ ≡ Then a2 + b2 = c2 . 6

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Theorem. If c is an integer such that c 2 , or 3 mod4 , ≡2 then the equation x = c has no integer solution. If both a and b are odd, then a2 + b2 1+1 2 mod4 . ≡ ≡ Then a2 + b2 = c2 . In a primitive Pythagorean triple (a,b,c) , 6 c is odd, one of a and b is even, another is odd.

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Theorem. If c is an integer such that c 2 , or 3 mod4 , ≡2 then the equation x = c has no integer solution. If both a and b are odd, then a2 + b2 1+1 2 mod4 . ≡ ≡ Then a2 + b2 = c2 . In a primitive Pythagorean triple (a,b,c) , 6 c is odd, one of a and b is even, another is odd. Without loss of generality, let us assume that a is odd and b is even.

6/11 Squares modulo 4

The square of an even number is divisible by four. n 0 mod2 = n2 0 mod4 ≡ ⇒ ≡ The square of an odd number has remainder 1 under division by 4 . n 1 mod2 = n2 1 mod4 ≡ ⇒ ≡ Notation: x y mod m means x y is divisible by m . ≡ − Theorem. If c is an integer such that c 2 , or 3 mod4 , ≡2 then the equation x = c has no integer solution. If both a and b are odd, then a2 + b2 1+1 2 mod4 . ≡ ≡ Then a2 + b2 = c2 . In a primitive Pythagorean triple (a,b,c) , 6 c is odd, one of a and b is even, another is odd. Without loss of generality, let us assume that a is odd and b is even. Rewrite a2 + b2 = c2 as b2 = c2 a2 − 2 and further as b =(c a)(c + a) . −

6/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡

7/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡ The sum and difference of two odd numbers is even .

7/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡ The sum and difference of two odd numbers is even . Denote c + a =2P , c a =2Q and b =2R . −

7/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡ The sum and difference of two odd numbers is even .

c + a = 2P c a = 2Q ( −

7/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡ The sum and difference of two odd numbers is even .

c + a = 2P c a = 2Q ( − This system can be solved as follows:

c = P + Q a = P Q ( −

7/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡ The sum and difference of two odd numbers is even .

c + a = 2P c a = 2Q ( − This system can be solved as follows:

c = P + Q a = P Q ( − For a primitive Pythagorean triple, P and Q have no common divisor.

7/11 New unknowns

c 1 mod2 c a 0 mod2 ≡ = − ≡ a 1 mod2 ⇒ c + a 0 mod2 ( ≡ ( ≡ The sum and difference of two odd numbers is even .

c + a = 2P c a = 2Q ( − This system can be solved as follows:

c = P + Q a = P Q ( − For a primitive Pythagorean triple, P and Q have no common divisor. Recall b2 = c2 a2 =(c + a)(c a)=4PQ =4R2 . − −

7/11 Solution

Summary: a primitive Pythagorean triple (a,b,c) can be obtained from relatively prime integers P and Q that are related to a,b,c by formulas c = P + Q , a = P Q and b2 =4PQ . −

8/11 Solution

Summary: a primitive Pythagorean triple (a,b,c) can be obtained from relatively prime integers P and Q that are related to a,b,c by formulas c = P + Q , a = P Q and b2 =4PQ . − Equation a2 + b2 = c2 is equivalent to PQ = R2 , where b =2R .

8/11 Solution

Summary: a primitive Pythagorean triple (a,b,c) can be obtained from relatively prime integers P and Q that are related to a,b,c by formulas c = P + Q , a = P Q and b2 =4PQ . − Equation a2 + b2 = c2 is equivalent to PQ = R2 , where b =2R .

P and Q are squares, since they are relatively prime and PQ is a square.

8/11 Solution

Summary: a primitive Pythagorean triple (a,b,c) can be obtained from relatively prime integers P and Q that are related to a,b,c by formulas c = P + Q , a = P Q and b2 =4PQ . − Equation a2 + b2 = c2 is equivalent to PQ = R2 , where b =2R .

P and Q are squares, since they are relatively prime and PQ is a square. P = u2 2 Q = v R = uv



8/11 Solution

Summary: a primitive Pythagorean triple (a,b,c) can be obtained from relatively prime integers P and Q that are related to a,b,c by formulas c = P + Q , a = P Q and b2 =4PQ . − Equation a2 + b2 = c2 is equivalent to PQ = R2 , where b =2R .

P and Q are squares, since they are relatively prime and PQ is a square. P = u2 a = u2 v2 2 − Q = v Express the Pythagorean triple: b = 2uv R = uv c = u2 + v2

 

8/11 Solution

Summary: a primitive Pythagorean triple (a,b,c) can be obtained from relatively prime integers P and Q that are related to a,b,c by formulas c = P + Q , a = P Q and b2 =4PQ . − Equation a2 + b2 = c2 is equivalent to PQ = R2 , where b =2R .

P and Q are squares, since they are relatively prime and PQ is a square. P = u2 a = u2 v2 2 − Q = v Express the Pythagorean triple: b = 2uv R = uv c = u2 + v2 Verify: for any integers u, v (no matter odd, relatively prime, or not)   a2 + b2 =(u2 v2)2 + (2uv)2 − = u4 2u2v2 + v4 +4u2v2 = u4 +2u2v2 + v4 − =(u2 + v2)2 = c2

8/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even.

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem.

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire?

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point?

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point? A mathematician would ask: What did we learn?

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point? A mathematician would ask: What did we learn? What other problems can we solve using this experience?

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point? A mathematician would ask: What did we learn? What other problems can we solve using this experience? What generalizations of this problem can be handled in this way?

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point? A mathematician would ask: What did we learn? What other problems can we solve using this experience? What generalizations of this problem can be handled in this way? It is not geometry.

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point? A mathematician would ask: What did we learn? What other problems can we solve using this experience? What generalizations of this problem can be handled in this way? It is not geometry. Until the proof has turned into a comics, some mathematicians would not be satisfied.

9/11 Done?

All primitive Pythagorean triples are (2uv,u2 v2,u2 + v2) for − relatively prime u and v , one of which is even. At this point we could stop: we have solved the problem. What else to desire? What may a mathematician desire at this point? A mathematician would ask: What did we learn? What other problems can we solve using this experience? What generalizations of this problem can be handled in this way? It is not geometry. Until the proof has turned into a comics, some mathematicians would not be satisfied. Pythagorean triples come from Geometry. Where is Geometry?

9/11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple.

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane.

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane. It has integer coordinates (a, b) and integer distance c = √a2 + b2 from the origin.

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane. It has integer coordinates (a, b) and integer distance c = √a2 + b2 from the origin. This is a geometric incarnation of Pythagorian triple (a,b,c) , a Pythagorian point.

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane. It has integer coordinates (a, b) and integer distance c = √a2 + b2 from the origin. This is a geometric incarnation of Pythagorian triple (a,b,c) , a Pythagorian point. If (a, b) is a Pythagorean point and c = √a2 + b2 , a b 2 2 then ( c , c ) is a rational point on the circle x + y =1 .

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane. It has integer coordinates (a, b) and integer distance c = √a2 + b2 from the origin. This is a geometric incarnation of Pythagorian triple (a,b,c) , a Pythagorian point. If (a, b) is a Pythagorean point and c = √a2 + b2 , a b 2 2 then ( c , c ) is a rational point on the circle x + y =1 . Conversly, if (p,q) is a rational point on x2 + y2 =1 , q then on the line y = p x there are Pythagorean points.

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane. It has integer coordinates (a, b) and integer distance c = √a2 + b2 from the origin. This is a geometric incarnation of Pythagorian triple (a,b,c) , a Pythagorian point. If (a, b) is a Pythagorean point and c = √a2 + b2 , a b 2 2 then ( c , c ) is a rational point on the circle x + y =1 . Conversly, if (p,q) is a rational point on x2 + y2 =1 , q then on the line y = p x there are Pythagorean points. The problem that we solved was find all Pythagorain points.

10 / 11 Pythagorian points

c = √a2 + b2 is recovered from the rest of Pythagorian triple. The rest is a point (a, b) on the Cartesian plane. It has integer coordinates (a, b) and integer distance c = √a2 + b2 from the origin. This is a geometric incarnation of Pythagorian triple (a,b,c) , a Pythagorian point. If (a, b) is a Pythagorean point and c = √a2 + b2 , a b 2 2 then ( c , c ) is a rational point on the circle x + y =1 . Conversly, if (p,q) is a rational point on x2 + y2 =1 , q then on the line y = p x there are Pythagorean points. The problem that we solved was find all Pythagorain points. Reformulation: find all rational points on the unit circle x2 + y2 =1 .

10 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α .

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈ Nice formulas! Homogeneous! Divide by u2

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈ Nice formulas! Homogeneous! Divide by u2 2 1− v u2 2v/u Then cos α = 2 and sin α = 2 . 1+ v 1+ v u2 u2

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈ Nice formulas! Homogeneous! Divide by u2 2 1− v u2 2v/u Then cos α = 2 and sin α = 2 . 1+ v 1+ v u2 u2 v 1−t2 2t Let u = t . Then cos α = 1+t2 and sin α = 1+t2

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈ Nice formulas! Homogeneous! Divide by u2 2 1− v u2 2v/u Then cos α = 2 and sin α = 2 . 1+ v 1+ v u2 u2 v 1−t2 2t Let u = t . Then cos α = 1+t2 and sin α = 1+t2 If u = cos β and v = sin β , then cos α = cos2 β sin2 β = cos2β , sin α = 2cos β sin β = sin 2β . −

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈ Nice formulas! Homogeneous! Divide by u2 2 1− v u2 2v/u Then cos α = 2 and sin α = 2 . 1+ v 1+ v u2 u2 v 1−t2 2t Let u = t . Then cos α = 1+t2 and sin α = 1+t2 If u = cos β and v = sin β , then cos α = cos2 β sin2 β = cos2β , sin α = 2cos β sin β = sin 2β . − v sin β α t = u = cos β = tan β = tan 2

11 / 11 Parametrizations of the circle

On the unit circle, any point is (cos α, sin α) for some α . Find α such that both cos α and sin α are rational.

u2−v2 2uv The answer: cos α = 2 2 and sin α = 2 2 for u, v N . u +v u +v ∈ Nice formulas! Homogeneous! Divide by u2 2 1− v u2 2v/u Then cos α = 2 and sin α = 2 . 1+ v 1+ v u2 u2 v 1−t2 2t Let u = t . Then cos α = 1+t2 and sin α = 1+t2 If u = cos β and v = sin β , then cos α = cos2 β sin2 β = cos2β , sin α = 2cos β sin β = sin 2β . − v sin β α t = u = cos β = tan β = tan 2 Riddle: Draw on a picture all the heroes: α , cos α , sin α , β , and t = tan β .

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